Conditional Probability

The Medical Test Dilemma

A medical test for a rare disease is 99% accurate. You test positive. Should you panic?

Surprising answer: It depends! If only 1 in 10,000 people have the disease, your actual probability of having it (even after testing positive) might be less than 10%!

This is conditional probability—how probabilities change when we have additional information. It’s crucial for JEE, medical diagnostics, weather forecasting, and machine learning!

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What is Conditional Probability?

Conditional probability is the probability of an event occurring, given that another event has already occurred.

Notation

P(A|B) reads as “probability of A given B”

It means: “What is the probability of A occurring, given that B has already occurred?”

Intuitive Example

Scenario: Drawing cards from a deck

• Original: P(drawing an ace) = 4/52
• Conditional: You’re told the card is red. Now P(ace | red card) = 2/26 = 1/13

The information “it’s red” reduces the sample space from 52 to 26 cards!

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Formula for Conditional Probability

Why This Formula?

When B occurs, the sample space effectively becomes B (not the full S). So:

• New sample space: B
• Favorable outcomes: A ∩ B
• Probability: P(A ∩ B) / P(B)

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Multiplication Theorem

Rearranging the conditional probability formula gives:

Use Case: When you want P(A ∩ B) but know P(A) and P(B|A)

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Independent Events

Definition

Two events A and B are independent if the occurrence of one does not affect the probability of the other.

Examples of Independence

Situation	Independent?	Why?
Two coin tosses	Yes	First toss doesn’t affect second
Drawing cards WITH replacement	Yes	First card is put back
Drawing cards WITHOUT replacement	No	First card changes remaining deck
Rain today and rain tomorrow	Maybe	Depends on weather patterns
Getting 6 on first die and second die	Yes	Dice don’t influence each other

Mutually Exclusive vs Independent

⚠️ Common Confusion:

• Mutually Exclusive: A ∩ B = ∅ (can’t both occur)
• Independent: P(A ∩ B) = P(A) × P(B)

Important: If A and B are mutually exclusive and both have non-zero probability, they cannot be independent!

Proof: If A ∩ B = ∅, then P(A ∩ B) = 0, but P(A) × P(B) ≠ 0, so P(A ∩ B) ≠ P(A) × P(B)

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Tree Diagrams

Tree diagrams help visualize sequential events and conditional probabilities.

How to Use Tree Diagrams

1. First branch: Draw outcomes of first event with probabilities
2. Second branch: For each first outcome, draw second event outcomes with conditional probabilities
3. Multiply along path: P(path) = product of probabilities along branches
4. Add paths: P(final outcome) = sum of all paths leading to it

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Properties of Conditional Probability

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Memory Tricks

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Common Mistakes to Avoid

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Solved Examples

Example 1: Basic Conditional Probability

Problem: A die is rolled. Find: (a) P(getting a 4 | even number) (b) P(even | number > 3)

Solution:

Sample Space: S = {1, 2, 3, 4, 5, 6}

(a) P(4 | even):

• Event A = {4}, Event B = {2, 4, 6} (even)
• A ∩ B = {4} $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{1/6}{3/6} = \frac{1}{3}$$

Alternative (reduced sample space): Given even, sample space = {2, 4, 6}, favorable = {4}

$$P(4|\text{even}) = \frac{1}{3}$$

(b) P(even | number > 3):

• Event A = {2, 4, 6}, Event B = {4, 5, 6}
• A ∩ B = {4, 6} $$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{2/6}{3/6} = \frac{2}{3}$$

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Example 2: Cards Without Replacement

Problem: Two cards are drawn from a deck without replacement. Find the probability that: (a) Both are aces (b) Both are red

Solution:

(a) Both aces:

$$P(\text{both aces}) = P(\text{1st ace}) \times P(\text{2nd ace | 1st ace})$$ $$= \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$$

(b) Both red:

$$P(\text{both red}) = P(\text{1st red}) \times P(\text{2nd red | 1st red})$$ $$= \frac{26}{52} \times \frac{25}{51} = \frac{650}{2652} = \frac{25}{102}$$

Note: Probabilities change because cards are not replaced!

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Example 3: Multiplication Theorem (Three Events)

Problem: A bag contains 5 red and 3 blue balls. Three balls are drawn without replacement. Find P(all three are red).

Solution:

$$P(\text{RRR}) = P(R_1) \times P(R_2|R_1) \times P(R_3|R_1 R_2)$$

• 1st draw: P(R₁) = 5/8
• 2nd draw (given 1st red, now 4 red, 3 blue, total 7): P(R₂|R₁) = 4/7
• 3rd draw (given 1st, 2nd red, now 3 red, 3 blue, total 6): P(R₃|R₁R₂) = 3/6 = 1/2

$$P(\text{RRR}) = \frac{5}{8} \times \frac{4}{7} \times \frac{1}{2} = \frac{20}{112} = \frac{5}{28}$$

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Example 4: Independence Check

Problem: Events A and B are such that P(A) = 0.6, P(B) = 0.5, P(A ∩ B) = 0.3. Are A and B independent?

Solution:

Check: Is P(A ∩ B) = P(A) × P(B)?

$$P(A) \times P(B) = 0.6 \times 0.5 = 0.3$$ $$P(A \cap B) = 0.3$$

Since $P(A \cap B) = P(A) \times P(B)$, Yes, A and B are independent.

Verification using conditional probability:

$$P(A|B) = \frac{P(A \cap B)}{P(B)} = \frac{0.3}{0.5} = 0.6 = P(A) \checkmark$$

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Example 5: Medical Test (JEE Advanced Pattern)

Problem: A disease affects 1% of the population. A test for the disease is 95% accurate (both sensitivity and specificity). If a person tests positive, what is the probability they actually have the disease?

Solution:

Let:

• D = person has disease
• T = test is positive

Given:

• P(D) = 0.01, so P(D’) = 0.99
• P(T|D) = 0.95 (true positive rate)
• P(T’|D’) = 0.95, so P(T|D’) = 0.05 (false positive rate)

Find: P(D|T)

Step 1: Find P(T) using total probability

$$P(T) = P(D) \times P(T|D) + P(D') \times P(T|D')$$ $$= 0.01 \times 0.95 + 0.99 \times 0.05$$ $$= 0.0095 + 0.0495 = 0.059$$

Step 2: Apply conditional probability

$$P(D|T) = \frac{P(D \cap T)}{P(T)} = \frac{P(D) \times P(T|D)}{P(T)}$$ $$= \frac{0.01 \times 0.95}{0.059} = \frac{0.0095}{0.059} \approx 0.161 = 16.1\%$$

Answer: Even with a positive test, only about 16% chance of actually having the disease!

Why? The disease is so rare that most positive tests are false positives.

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Example 6: Urn Problem (Classic JEE)

Problem: Urn A contains 3 red and 2 blue balls. Urn B contains 2 red and 4 blue balls. An urn is selected at random and a ball is drawn. Find: (a) P(red ball) (b) If the ball is red, P(it came from urn A)

Solution:

(a) P(red):

$$P(R) = P(A) \times P(R|A) + P(B) \times P(R|B)$$ $$= \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{2}{6}$$ $$= \frac{1}{2} \times \frac{3}{5} + \frac{1}{2} \times \frac{1}{3}$$ $$= \frac{3}{10} + \frac{1}{6} = \frac{9 + 5}{30} = \frac{14}{30} = \frac{7}{15}$$

(b) P(A|R):

$$P(A|R) = \frac{P(A \cap R)}{P(R)} = \frac{P(A) \times P(R|A)}{P(R)}$$ $$= \frac{\frac{1}{2} \times \frac{3}{5}}{\frac{7}{15}} = \frac{\frac{3}{10}}{\frac{7}{15}} = \frac{3}{10} \times \frac{15}{7} = \frac{45}{70} = \frac{9}{14}$$

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Practice Problems

Level 1: Foundation (JEE Main)

1. A bag has 4 red and 6 blue balls. Two balls are drawn without replacement. Find P(both blue).

2. P(A) = 0.4, P(B) = 0.5, P(A ∩ B) = 0.2. Find P(A|B).

3. A die is rolled. Given the number is odd, find P(number is prime).

Solutions

1. P(both blue) = P(B₁) × P(B₂|B₁) = (6/10) × (5/9) = 30/90 = 1/3

2. P(A|B) = P(A ∩ B)/P(B) = 0.2/0.5 = 0.4

3. Odd = {1, 3, 5}, Prime among odd = {3, 5} P(prime|odd) = 2/3

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Level 2: Intermediate (JEE Main/Advanced)

4. Two dice are thrown. Given the sum is 7, find P(at least one die shows 3).

5. A speaks truth 60% of the time, B speaks truth 80% of the time. They both describe an event. Find P(they contradict each other).

6. Three cards are drawn from a deck without replacement. Find P(all are face cards).

Solutions

4. Sum = 7: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}, total 6 outcomes At least one 3: {(3,4), (4,3)}, total 2 outcomes P = 2/6 = 1/3

5. P(contradict) = P(A true, B false) + P(A false, B true) = 0.6 × 0.2 + 0.4 × 0.8 = 0.12 + 0.32 = 0.44

6. Face cards = 12 in deck P = (12/52) × (11/51) × (10/50) = 1320/132600 = 11/1105

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Level 3: Advanced (JEE Advanced)

7. Bag I: 3 red, 2 blue. Bag II: 2 red, 3 blue. A die is thrown. If 1 or 2 appears, a ball is drawn from Bag I; otherwise from Bag II. Find P(ball is red).

8. Events A, B, C are such that P(A) = P(B) = P(C) = 1/4, P(A ∩ B) = P(B ∩ C) = 0, P(A ∩ C) = 1/8. Find P(C | A ∪ B).

9. Prove: If P(A|B) > P(A), then P(B|A) > P(B).

Solutions

7. P(Bag I) = 2/6 = 1/3, P(Bag II) = 4/6 = 2/3 P(R|I) = 3/5, P(R|II) = 2/5 P(R) = (1/3)(3/5) + (2/3)(2/5) = 1/5 + 4/15 = 3/15 + 4/15 = 7/15

8. P(A ∪ B) = P(A) + P(B) - P(A ∩ B) = 1/4 + 1/4 - 0 = 1/2 P(C ∩ (A ∪ B)) = P((C ∩ A) ∪ (C ∩ B)) = P(C ∩ A) + P(C ∩ B) - P(C ∩ A ∩ B) = 1/8 + 0 - 0 = 1/8 P(C|A ∪ B) = (1/8)/(1/2) = 1/4

9. P(A|B) > P(A) → P(A ∩ B)/P(B) > P(A) → P(A ∩ B) > P(A)P(B) P(B|A) = P(A ∩ B)/P(A) > P(A)P(B)/P(A) = P(B) ✓

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Real-Life Applications

1. Medical Diagnosis

• P(disease | positive test) ≠ P(positive test | disease)
• Need to account for base rate (disease prevalence)
• False positives more common when disease is rare

2. Weather Forecasting

• P(rain tomorrow | rain today) > P(rain tomorrow)
• Conditional on current atmospheric conditions
• Bayesian updating with new data

3. Spam Filtering

• P(spam | word) calculated for each word
• Combines multiple conditional probabilities
• Naive Bayes classifier

4. Sports Analytics

• P(win | specific player plays) vs P(win)
• Conditional on opponent, home/away, etc.
• Strategy optimization

5. Card Games (Poker, Blackjack)

• Probabilities change as cards are revealed
• Counting cards = tracking conditional probabilities
• Optimal strategy based on seen cards

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Connection to Other Topics

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Formula Quick Reference

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Quick Revision Checklist

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Exam Tips

1. Identify “given”: P(A|B) means “given B occurred”
2. Check independence: If independent, skip conditional formula and just multiply
3. Draw tree diagrams: For sequential events, visualize with trees
4. Without replacement: Remember to reduce numerator AND denominator
5. Read carefully: P(A|B) ≠ P(B|A) in general
6. Use complement: Sometimes P(A’|B) is easier to find than P(A|B)

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Next Steps

Ready to flip probabilities around? Continue to:

• Bayes’ Theorem - The ultimate tool for reversing conditional probabilities

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Last updated: December 18, 2025 Practice conditional probability at JEENotes Practice Portal